数学（文科）-内蒙古赤峰市2024届高三下学期一模模拟考试试题和答案

1当即2024.03赤峰市高三年级3.20当即2024.03文科数学答案个选项中，只有一项是符合题目要求的.123456789CBACBDADBBCDSn−1nnn−1，............................................................ =2...................................所以an=2n................................................................................................................................6nn2222..................................................................n(n−1)22......................................................................................=2符合上式..............................所以an=2n................................................................................................................................62因为bn=log2an，所以bn=n，.................(1)(11)(11) 因为neN*所以Tn<1 (0.003+0.005+0.012)x10∴对甲商场评分低于30的人数为100x0.2=20人，......从这5人中随机选出2人的选法为：甲商场评分低于30的人数为20，∴甲商场评分低于30分的人数所占的比例为20%，.∴乙商场得分低于30分的人数所占的比例为10%，.........∴会选择乙商场消费..................................................（ii x甲=10x（5x0.003+15x0.005+25x0.012+35x0.020+45x0.020+55x0.040）3=41.9................................................xx甲乙∴会选择乙商场消费。...............xx则函数g(x)在(0,1)上有一个异号零点，.............................24法II：令(a−1)x2+x−1=0.................................................................................................7分2 xx 所以A1C⊥B1D1..............................................................（2）截面图形，如图所示的六边形............................................................................所以最大的截面面积为S=6xxxxsin60所以S=3..............................................................................................................易求SΔFGC=，因为点E到平面FGC的距离为1，53..........................................................................所以SΔEGC=设F到平面EGC的距离为dΔEGC 所以F到平面EGC的距离为......................................................................................12分2(p)=4x......................................................（2）设直线AB方程为x=my+1,且A(x1,y1)，B(x2,y2),..................................................5分2则y1+y2=4m,y1y2=−4,........则AB=.................==4m2+4.................................................. m2，..................................................................................................[注]：本题如果直接使用过焦点的弦长公式AB=解答，本问分数减半；如果有此(x=cosθ(x=cosθly=4+sinθly−4=sinθ6所以曲线C2的普通方程为x2+(y−4)2=1，又因为2..........所以曲线C2的普通方程为x2+(y−4)2=1（3y4）.........................................................4分y=sin所以设点M的坐标为(2cos,sin)，..............................................................................5分设圆心C2与C1上任意一点的距离为d则d=..............................................................................=..................................................................................=1919−248..............................................................9分dmin当x−1时，不等式化为−x+2−x−14，解得x−，所以x−.......................2分52当x2时，不等式化为x−2+x+14，解得x5252,所以x523535②若f(x)2m−x+1恒成立，即x−m+x+1min2m................................................6分因为x−m+x+1x−m−x−1=m+1(当且仅当(x−m)(x+1)0时，等号成立)..7分所以m+12m，即m+12m或m+1−2m，..................................................................8分解得m1或m